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3 years ago

How much work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s?

Physics

1 answer:

soldi70 [24.7K]3 years ago

6 0

Answer:

The correct option is (b).

Explanation:

We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.

We know that, the work done is equal to the kinetic energy of an object i.e.

$W=\Delta K\\\\W=\dfrac{1}{2}mv^2\\\\W=\dfrac{1}{2}\times 1\times 5^2\\W=12.5\ J$

So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.

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If the speed of an object doubles, how does that affect its kinetic energy? A. Halves B. Doubles C. Quarters D. Quadruples

kvasek [131]

Answer is :

D. Quadruples

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3 years ago

Read 2 more answers

A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at

Readme [11.4K]

Answer:

KE = 1.75 J

Explanation:

given,

mass of ball, m₁ = 300 g = 0.3 Kg

mass of ball 2, m₂ = 600 g = 0.6 Kg

length of the rod = 40 cm = 0.4 m

Angular speed = 100 rpm= $100\times \dfrac{2\pi}{60}$

=10.47\ rad/s

now, finding the position of center of mass of the system

r₁ + r₂ = 0.4 m.....(1)

equating momentum about center of mass

m₁r₁ = m₂ r₂

0.3 x r₁ = 0.6 r₂

r₁ = 2 r₂

Putting value in equation 1

2 r₂ + r₂ = 0.4

r₂ = 0.4/3

r₁ = 0.8/3

now, calculation of rotational energy

$KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2$

$KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)$

$KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)$

$KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)$

KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J

7 0

3 years ago

Question 5 of 10

muminat

The correct answer to this question is D

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3 years ago

Chi walked 100 meters in 5 minutes. She took another 15 minutes to walk another 400 meters to reach her house. What is her avera

Law Incorporation [45]

Average speed is the ratio of total distance moved by Chi in total time interval

So here we will have

Total distance = 100 m + 400 m

$d = 500 m = 0.5 km$

Total time taken = 5 min + 15 min = 20 min

$T = \frac{20}{60} = \frac{1}{3} hour$

now by the formula of average speed we know that

$v_{avg} = \frac{d}{T}$

$v_{avg} = \frac{0.5 km}{1/3 h}$

$v_{avg} = 1.5 km/h$

so average speed will be 1.5 km/h

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iVinArrow [24]

Answer:

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